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Rupert, I don't understand your proof. It might take a bit of work for
you to explain it to me, so I understand that such work may not be
enjoyable for you; but if, on the the other hand, you don't mind
working it through with me, I do thank you for your time and effort.
Rupert wrote:
> Suppose bXc=b and b!=0. Let x be in b.
Okay, I take that as existential instantiation to the variable x. I
only belabor that point because it bears upon why I don't understand
the rest of the proof.
> Say that yRx if there exists a z
> in c such that x=<y,z>, or there exists a w in b and a z in c such that
> y={w} and x=<w,z>.
Is x here under a universal quantifier or is it unquantified so that it
is the x we said is in b?
If x is under a universal quantifier, then I see that R is defined as a
predicate (but not yet proven to be a set) for all x and y, but without
any mention of the x that we said is in b. But if x is the x we said is
in b, then isn't R a 1-place predicate thus not a relation?
> Even without infinity, we can define the class of
> objects that bear the transitive closure of R to x, though we cannot
> prove it is a set. However, since it is contained in b union P(b),
I know now what the transitive closure of a relation on a set is. So
you mean that the transitive closure of R is a subset of buPb? But I'm
still confused about what R is and what set it is a relation on, so I
don't know how to see it's a subset of buPb.
> it
> is a set. Now we can easily show that it has no minimal element,
> contradiction.
Thanks,
MoeBlee
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